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Unique behaviour of the first element in each group: NEET MCQs with answers

50 questions on this topic of P-Block Elements (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Consider the hydrolytic behaviour of tetrahalides of Group 14. While SiCl₄ hydrolyses readily in water, CCl₄ is completely inert toward hydrolysis under ambient conditions. What is the fundamental reason for this distinction?

  1. C-Cl bond is completely non-polar compared to Si-Cl bond
  2. Carbon lacks vacant d-orbitals to accept a lone pair from attacking water molecules
  3. Carbon has a higher electronegativity than silicon
  4. CCl₄ is a planar molecule while SiCl₄ is tetrahedral
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Answer: (B) Carbon lacks vacant d-orbitals to accept a lone pair from attacking water molecules

Hydrolysis of group 14 tetrahalides proceeds via coordination of a water molecule to the central atom. Silicon can expand its octet by accepting lone pairs into vacant 3d-orbitals, whereas carbon possesses only 2s and 2p valence orbitals and cannot coordinate with water.

medium

2. Nitrogen cannot form NCl₅, but phosphorus forms PCl₅, because nitrogen:

  1. Has no vacant d orbitals, so its maximum covalency is 4
  2. Is more metallic
  3. Has a larger size
  4. Has fewer valence electrons
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Answer: (A) Has no vacant d orbitals, so its maximum covalency is 4

First members of p-block groups lack d orbitals in their valence shell, limiting maximum covalency to 4. Phosphorus can use 3d orbitals to expand its octet.

NCERT Class 11

hard

3. Match the anomalous feature of the first-row p-block element in List-I with the corresponding species in List-II: List-I P. Octet expansion is prohibited Q. Unusually high standard reduction potential for F₂/F⁻ R. Only stable gaseous diatomic molecule with a triple bond S. Polymerises into multicentre electron-deficient hydride List-II 1. B₂H₆ 2. N₂ 3. NF₃ (inability to form NF₅) 4. High hydration enthalpy of F⁻ combined with low F-F bond energy

  1. P-4, Q-3, R-1, S-2
  2. P-3, Q-4, R-2, S-1
  3. P-3, Q-1, R-2, S-4
  4. P-2, Q-4, R-3, S-1
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Answer: (B) P-3, Q-4, R-2, S-1

P matches 3 because nitrogen cannot expand its octet to form NF₅. Q matches 4 because the high reduction potential of F₂ is due to low dissociation energy and very high hydration enthalpy of small F⁻. R matches 2 because N₂ forms a stable triple bond. S matches 1 because boron forms diborane B₂H₆ via multicentre bonding.

medium

4. Boron forms the monomeric planar halide BF₃, which acts as a Lewis acid by accepting an electron pair. Which factor explains why BF₃ is a weaker Lewis acid than BCl₃ and BBr₃?

  1. Effective 2pπ-2pπ back-bonding between boron and fluorine orbitals
  2. Greater steric hindrance offered by smaller fluorine atoms
  3. Higher electronegativity of fluorine polarizing the B-F bond
  4. Complete octet completion in BF₃ by dimer formation
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Answer: (A) Effective 2pπ-2pπ back-bonding between boron and fluorine orbitals

Due to the comparable size of 2p orbitals of boron and fluorine, effective pπ-pπ back-bonding occurs, partially compensating for the electron deficiency of the boron atom. In BCl₃ and BBr₃, overlap between 2p of B and 3p/4p of halogens is inefficient, making them stronger Lewis acids.

easy

5. Why does oxygen form a neutral diatomic gaseous molecule (O₂) at room temperature, while sulfur exists as an octatomic puckered ring solid (S₈)?

  1. Sulfur has a higher electronegativity than oxygen, favouring cyclic oligomerization
  2. Oxygen forms strong pπ-pπ double bonds due to its small atomic size, whereas sulfur prefers single bonds
  3. The vacant d-orbitals in oxygen force it to adopt an sp² hybridization state
  4. Oxygen has a lower ionization enthalpy, promoting rapid dissociation into atoms
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Answer: (B) Oxygen forms strong pπ-pπ double bonds due to its small atomic size, whereas sulfur prefers single bonds

Due to its very small size, oxygen effectively forms strong pπ-pπ multiple bonds to produce stable O=O molecules. Larger elements like sulfur have diffuse p-orbitals that yield poor lateral overlap, making the formation of eight single S-S bonds in an S₈ ring energetically more favourable.

hard

6. Given below are two statements: Statement I: Carbon forms stable homonuclear triple bonds (C≡C) in alkynes, whereas silicon does not readily form stable, unhindered Si≡Si triple-bonded compounds. Statement II: The large, diffuse 3p orbitals of silicon cannot overlap laterally with sufficient efficiency to form strong π-bonds. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
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Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

The small covalent radius of carbon allows its 2p orbitals to approach closely and achieve effective lateral overlap to form strong pπ-pπ multiple bonds. In contrast, silicon's 3p orbitals are larger and diffuse, resulting in weak sideways overlap that makes multiple bonds unstable under normal conditions.

easy

7. Given below are two statements: Statement I: Nitrogen shows a maximum covalency of 4 in its compounds. Statement II: Nitrogen has only four valence orbitals (one 2s and three 2p) available for bonding.

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

The first member of group 15, nitrogen, has only four valence orbitals (one 2s and three 2p) and lacks d-orbitals in the second energy level. Consequently, its maximum covalency is restricted to 4, as seen in the ammonium ion [NH₄]⁺.

easy

8. Among the Group 15 elements, bismuth forms BiF₅ as its only stable pentahalide, whereas nitrogen does not form NF₅ at all. What is the fundamental electronic restriction preventing nitrogen from forming NF₅?

  1. Unfavourable standard reduction potential of nitrogen
  2. High electronegativity difference between nitrogen and fluorine
  3. Absence of d-orbitals in the valence shell of nitrogen
  4. Exceptionally high third ionization enthalpy of nitrogen
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Answer: (C) Absence of d-orbitals in the valence shell of nitrogen

Nitrogen belongs to the second period and possesses only 2s and 2p valence orbitals, with no accessible 2d orbitals. Consequently, its maximum covalency is restricted to 4, completely preventing the formation of pentavalent halides like NF₅.

easy

9. Fluorine shows only the -1 oxidation state because it:

  1. Has a large size
  2. Has seven unpaired electrons
  3. Is the most electronegative element and has no d orbitals
  4. Is a metal
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Answer: (C) Is the most electronegative element and has no d orbitals

Fluorine is the most electronegative element and lacks d orbitals, so it cannot show positive oxidation states; other halogens show +1 to +7.

NCERT Class 11

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10. Which of the following compounds of group 15 exhibits planar geometry around the central heteroatom due to back-bonding involving a second-period donor atom and a third-period acceptor atom?

  1. (SiH₃)₃N
  2. NF₃
  3. (CH₃)₃N
  4. N(CH₃)₂Cl
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Answer: (A) (SiH₃)₃N

In trisilylamine, (SiH₃)₃N, the lone pair of electrons on nitrogen (second-period atom) is transferred into the empty 3d-orbitals of adjacent silicon atoms via pπ-dπ back-bonding. This interaction changes nitrogen's hybridization from sp³ to sp², giving a completely planar geometry.

easy

11. Which of the following group 14 elements exhibits the highest tendency for catenation?

  1. Carbon
  2. Tin
  3. Silicon
  4. Lead
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Answer: (A) Carbon

Carbon has a unique and exceptionally high tendency for catenation compared to other group 14 elements due to its small atomic size and very high C-C single bond enthalpy (348 kJ mol⁻¹). Down the group, atomic size increases and M-M bond enthalpy decreases sharply, causing catenation tendency to decrease.

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12. Consider the following statements regarding group 13 halides: Statement I: BF₃ is a monomeric gas, whereas AlCl₃ exists as a stable dimer, Al₂Cl₆, in non-polar solvents. Statement II: Boron cannot form halide bridges because its small radius cannot accommodate four large halogen atoms without severe steric repulsion, and it relieves electron deficiency via internal pπ-pπ back-bonding. Choose the correct option:

  1. Statement I is correct, but Statement II is incorrect
  2. Statement I is incorrect, but Statement II is correct
  3. Both Statement I and Statement II are correct
  4. Both Statement I and Statement II are incorrect
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Answer: (C) Both Statement I and Statement II are correct

BF₃ achieves partial stability through efficient 2pπ-2pπ back-bonding between boron and fluorine and does not dimerize because boron is too small to coordinate four fluorine atoms effectively. AlCl₃ cannot undergo efficient back-bonding and readily forms chlorine-bridged dimers (Al₂Cl₆) to complete its octet.

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